How Do You Calculate Displacement and Acceleration in a Kinematics Car Problem?

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The discussion focuses on calculating displacement and acceleration for a car problem involving three phases of motion. After accelerating at 3.0 m/s² for 5.0 seconds, the car reaches a velocity of 15 m/s. During the braking phase, the calculated acceleration is corrected to -1.9 m/s². The total displacement is initially calculated as 403.1 meters, but adjustments are needed based on the corrected acceleration value. Accurate calculations are essential for determining the correct displacement and acceleration in kinematics problems.
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A car starts from rest heading due east. It first accelerates at 3.0m/s^2 for 5.0s and then continues without further acceleration for 20.0s. It then brakes for 8.0s in coming to rest. (a) What is the car's velocity after the first 5.0 seconds? (b) What is the car's acceleration over the last 8.0s interval? (c) What is the total displacement?






The Attempt at a Solution



This is what i got: a) vf=vi+at
=0(3)(5)
vf=15m/s

b) vf=vi+at
0=15+a(8)
-15=a(8)
a=-1.7m/s^2

c) d1=37.5m
d2=300m
d3=65.6m
dtot=403.1m
 
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miamiheat5 said:
This is what i got: a) vf=vi+at
=0(3)(5)
vf=15m/s

Correct.

b) vf=vi+at
0=15+a(8)
-15=a(8)
a=-1.7m/s^2
15/8 is not 1.7

c) d1=37.5m
d2=300m
d3=65.6m
dtot=403.1m

d3 would change based on the answer to b.
 
ya it should be -1.9
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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